<p>Can anyone please answer these and leave an explanation? </p>
<p>I found these really hard...</p>
<p><a href="http://i.imgur.com/3IPtS.png%5B/url%5D">http://i.imgur.com/3IPtS.png</a></p>
<p><a href="http://i.imgur.com/NgoQV.png%5B/url%5D">http://i.imgur.com/NgoQV.png</a></p>
<p>For number 15, I just used my CAS function of my calculator to solve 3 different systems of equations... I don't know how I could have done this problem without doing this.</p>
<p>^^ </p>
<p>Try differences of squares. What is (p+n)(p-n)= ?</p>
<p>Number 19 is a factoring problem. Factor out a (2x)^y. The first expression [(2x)^3y] becomes (2x)^y. And the second expression [(2x)^y] becomes 1.</p>
<p>For #20, j and n are 2 digits apart, like 1 and 3 or 2 and 4. So what 2 digits like this result in a product with 9 as the units digit?</p>
<p>1 x 3 = 3 No
2 x 4 = 8 No
3 x 5 = 15 No
ETC
9 x 11 = 99 YES</p>
<p>So if j = 9 and n = 11, then k = 10. The units digit is zero.</p>
<p>Xiggi, okay, I see that now. But, how would I use it to solve the problem? </p>
<p>(p+n)(p-n) = 12</p>
<p>? </p>
<p>Do I somehow test all of the answer choices? Or do test all of the factors of 12?</p>
<p>Essay, wow, I can’t believe I overlooked that for 19… Thanks! </p>
<p>And okay, for problems like 20, is the best way really to test out values? Gahh, I keep reverting back to the conventional, school method for doing math. I know that that’s bad…</p>
<p>Doing #20 in my head took all of about 20 seconds. I definitely think it’s faster. That’s why the SAT is so hard, though—learning to abandon the methods that have been forced on us for so long is tough to do. But I rarely rely on traditional math for questions 18, 19, and 20. There is usually a shortcut.</p>
<p>Number 15 can be very easy if you know how to look at it. When it says p^2 - q^2 = 12, all it is saying is that the difference between the squares is 12, so what perfect squares are 12 apart? If you just think about the first few perfect squares, 1, 4, 9, 16, 25… That’s all you need to go up to, because 16 - 4 = 12, and that is the only one that works. So p = sqrt(16) = 4 and q = sqrt(4) = 2. 4 - 2 = 2, which is your only answer.
Sorry this is such a long/complicated answer, I just wanted to make sure you got my thought process, but if you see it like that it only took me about 15 seconds to get the answer.</p>
<p>Oops, I meant n, not q!</p>
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<p>Test the factors that yield possible answers:</p>
<p>(p+n)(p-n) = 12
12 * 1
6 * 2
3 * 4 </p>
<p>Only 6*2 works. You’re done.</p>
<p>Xiggi, I’m trying to see how you’re doing the problem but I still don’t get it entirely. </p>
<p>Why do you say only 6*2 works? </p>
<p>The answer is B so p-n can only equal 2. </p>
<p>But, what about 3*4? Why can’t p-n = 4?</p>
<p>Did you mentally choose 2 numbers to add together and also subtract together to get 6 and 2? Like, You choose 4 and 2. 4+2 = 6 and 4-2 = 2?</p>
<p>12 can be factored as 12<em>1, 6</em>2, 4*3, and that’s it (as a product of exactly 2 factors). </p>
<p>If p-n=4, then p+n=3. Adding, gives 2p=7. So p isn’t an integer.</p>
<p>From this example it’s easy to see that p+n and p-n must have the same parity. So only 6 and 2 CAN work. And you already showed that it DOES work.</p>
<p>Xiggi’s train of thought works as well. It couldn’t be 3<em>4 because p and n are both positive integers, and you can’t have 2 integers where p+n=4 and p-n=3. he just looked at the 3 possible ways to multiply 2 factors to equal 12: 1</em>12, 2<em>6, and 3</em>4, and the only way that works with 2 positive integers p and n is 2<em>6, and since the problem was (p+n)</em>(p-n)=12, (p-n) must equal 2, which is your answer. Again sorry for the very long response!</p>
<p>Oh, okay. Thanks so much for the explanation!!</p>
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<p>FYI, it is always important to pay close attention to the problem statement. One of the first clues is that p > n > 0, meaning that p and n are positive and that p+n has to be larger than p-n. Hence p+n cannot be smaller than 4.</p>
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<p>No. Actually, I did not pay attention to what p or n might be. The question is about p-n … not p or n. The question is also about testing p-n being equal to 1, 2, or 4. So, by isolating p-n (through the differences of squares) you focus immediately on the possible answers. </p>
<p>Again, remember that the SAT does not care for superfluous data. Just what leads to the answer. It is NOT important to know p or n individually.</p>
<p>But xiggi, the option could also be 4*3, making (p+n)=4 and (p-n)=3, which satisfies the given conditions. You can’t cancel it out until you realize that there are no 2 digits that when added together and subtracted end up as consecutive integers.</p>
<p>^^</p>
<p>Nope. The question only offers THREE values for p-n and none of them is 3. You are asked to test the proposed values of 1, 2, 4. Testing p-n = 3 (as in 4*3) is irrelevant here.</p>
<p>Yes, that also works. I was simply stating the point that you can’t discredit 3*4 only because p cannot be less than 4, as you said before.</p>
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<p>It is not a matter of “also working.” The point you are trying to make is not one … I made, namely that p cannot be less than 4. p+n is not the same as p.</p>
<p>Ok, sorry I think I explained myself incorrectly. before, you said that because p>n>0, p+n must be greater than p-n, and so p+n cannot be less than 4, which is all true. But you used that as proof that 3<em>4 cannot be considered. I was just trying to clarify that your reasoning there is not the actual reasoning to disregard 3</em>4 when doing the problem, but the OP has to also realize at least one of the other facts that both you and I have stated.</p>
